Problem 1.
The following matrix A is diagonalizable.
method.

−3 −8

A=
 −4 −8
−4 −10
Problem 2.
The following matrix is nilpotent.

0
0
N =
0
0
Find eAt by the diagonalization

10

11 

13
Find etN .

1 0 0
0 1 0
.
0 0 1
0 0 0
Problem 3.
The following matrix J is in Jordan Canonical Form. Find the matrices S
and N so that J = S + N , S is diagonalizable (diagonal, in this case), N is
nilpotent, and SN = N S. Compute eJt = eSt eN t by computing etS , etN and
multiplying.


1 1 0 0 0 0


 0 1 1 0 0 0 




 0 0 1 0 0 0 


J =

 0 0 0 2 1 0 




 0 0 0 0 2 0 


0
0
0
1
0
0
3
Problem 4.
Consider the matrix


568 −347 −31
1257 −1079 −215


 791 −485 −61
1767 −1504 −312 





 −681
419
45
−1517
1299
262


A=
.

 −620
393
37
−1393
1209
229





 −686
432
46
−1542
1331
259


129 −60 −6
261 −204 −59
A. Find a basis of each of the generalized eigenspaces of A.
B. Find the matrices S and N so that A = S + N , S is diagonalizable, N
is nilpotent, and SN = N S. Verify that your answers for S and N really
satisfy these conditions. (It is not necessary to go all the way to the Jordan
Canonical Form of A.)
C. Use S and N to compute eAt = etS etN .
2
Problem 5.
In each part, use Leonard’s Algorithm to find eAt .
A.

5 −1

A=
 12
−2
3 −1
B.

0


0 

2
1 1 −2




A=
 −3 5 −6 
0 1 −1
C.

1 1 −2




−3
5
−6
A=


0 1 −1
D.

−79 −4 −36

 614
7

A=
 262
4

71 −2
76


−558 


137 −240 

43 −61
325
Problem 6. Recall that we have shown
et(A+B) = etA etB
provided that A and B commute. To see what happens if A and B do not
commute, consider the function
Ψ(t) = et(A+B) e−tB e−tA ,
so if A and B commute, Ψ(t) = I for all t. To see how much Ψ differs from
the identity in the general case, find the first two nonzero terms in the Taylor
expansion of Ψ(t), centered at t = 0.
Problem 7.
Consider the following mechanical system.
3
We have two weights of masses m1 and m2 connected by three springs to a
stationary frame. The spring constants of the three springs are k1 , k2 and k3 , as
indicated in the diagram. The weights are constrained to move in a horizontal
line. The deviation of the two weights from their equilibrium positions are y1
and y2 respectively, measured positive to the right.
A. Assuming there is no friction, write down the system of linear equations that
governs the motion of the weights, using Hooke’s law for the force produced
by the springs. By very careful with your signs! Make this system into a first
order system by introducing new dependent variables z1 and z2 and adding
the equations y10 = z1 and y20 = z2 (see Section 3.1 in the book). Write the
resulting system in matrix form dy/dt = Ay.
B. Assume the values
m1 = 1,
m2 = 2,
k1 = 1,
k2 = 2,
k3 = 2.
Find eAt , where A is the coefficient matrix of the system, using one of the
methods developed in class.
Find the solution for the initial conditions
y1 (0) = 1,
y2 (0) = −1,
y10 (0) = 0,
y20 (0) = 0.
Provide a graph of y1 and y2 as functions of time.
C. Assume now that the weights are subject to frictional forces proportional to
their velocities, i.e., the force on the first weight due to friction is −b1 y10 and
the force on the second weight due to friction is −b2 y20 , where b1 and b2 are
positive constants. Find the new system of equations dy/dt = By.
D. Assume the values
m1 = 1, m2 = 1
k1 = 1, k2 = 1, k3 = 1
b1 = 1/2, b2 = 1/2.
Find eBt , using one of the methods developed in class. Solve the system
subject to the initial conditions
y1 (0) = 1,
y2 (0) = 0,
y10 (0) = 0 y20 (0) = 0.
Provide a graph showing y1 and y2 as functions of time.
4
Problem 8.
Do Problem 20 on page 190 of the book. See Section 1.7 in the book for how
to model electric circuits.
5
EXAM
Practice Exam # 2
Math 3351, Spring 2003
March 30, 2003
• Write all of your answers on separate sheets of paper.
You can keep the exam questions when you leave.
You may leave when finished.
• You must show enough work to justify your answers.
Unless otherwise instructed,
give exact answers, not
√
approximations (e.g., 2, not 1.414).
• This exam has 8 problems. There are 0 points
total.
Good luck!